Let us consider OAB be the equilateral triangle inscribed in parabola y^{2} = 4ax.

Let AB intersect the x – axis at point C.

Diagrammatic representation of the ellipse is as follows:

Now let OC = k

From the equation of the given parabola, we have,

So, y^{2} = 4ak

y =** ±**2\( \sqrt{ak}\)

The coordinates of points A and B are (k, \( 2 \sqrt{ak}\)), and (k, \( -2 \sqrt{ak}\))

AB = CA + CB

= \( 2 \sqrt{ak}+ 2 \sqrt{ak}\)

= \( 4\sqrt{ak}\)

Since, OAB is an equilateral triangle, OA^{2} = AB^{2}.

Then,

k^{2} + (\( 2 \sqrt{ak}\))^{2} = (\( 4 \sqrt{ak}\))^{2}

k^{2} + 4ak = 16ak

k^{2} = 12ak

k = 12a

Thus, AB = \( 4\sqrt{ak}\)

= \( 4\sqrt{(a×12a)}\)

= \(4\sqrt{12a^2}\)

= \( 4\sqrt{(4a×3a)}\)

= 4(2)\( \sqrt{3a}\)

= \( 8\sqrt{3a}\)

Hence, the side of the equilateral triangle inscribed in parabola

y^{2} = 4ax is \( 8\sqrt{3a}\).

An equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus rectum.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.